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Chapter 02 - Problems in Interest
Interest problems involve four quantities:
(a) Principal Invested
(b) Length of Investment (c) Rate of Interest (Discount) (d) Amount at the End of the Investment Period
Interest problems typically require solution for one or more of these variables, using values that are given or derived for the others.
Section 2.3 - Equations of Value
In these problem settings two equivalent investments are described in which one of the above quantities is unknown. Two valuation formulas are created (one for each investment) relative to some common comparison date. The two are equated to produce one equation in one unknown, an equation of value.
When using compound interest, the answer on the If simple interest is used, the answer
A Time Diagram graphically displays each payment (money coming in and money going out) at its appropriate time point and projects each payment to one common time point (see examples to follow). It is a very useful device to set up an equation of value.
Example: To receive a loan of $500 today and another $500 five years from today, you agree to pay $300 at the end of years 3, 6, 9, and the remainder at the end of year 12. What is the final payment if interest is accrued at a nominal interest rate of 5% convertible semiannually?
Then solving for x produces
x =
500 + 500 ν^10 − 300 ν^6 − 300 ν^12 − 300 ν^18 ν^24 = $ 391. 58.
Now repeat this problem solution using t = 5 as the comparison time point.
Time
Payment
Now each payment is accumulated or discounted to t = 5. Again set the present values of the two payment streams equal to one another producing
with i = .025. Then solving for x produces
x =
500 ν−^10 + 500 − 300 ν−^4 − 300 ν^2 − 300 ν^8 ν^14
If both the numerator and denominator are multiplied by ν^10 , the expression is exactly the same expression as the solution on the previous page. So the answer is the same, namely $391..
ν =
d(^4 ) 4
We use t = 1 as the vision point and form:
100 ( 1. 025 ) + 100 ( 1. 025 )^3 ν^8 + 100 ( 1. 025 )^5 ν^16 = sν^24
s = 100 ( 1. 025 ) + 100 ( 1. 025 )^3 ν^8 + 100 ( 1. 025 )^5 ν^16 ν^24 = $ 483. 11
Exercise 2-2:
You have an inactive credit card with a $1000 outstanding unpaid balance. This particular card charges interest at a rate of 18% compounded monthly. You are able to make a payment of $200 one month from today and $300 two months from today. Find the amount you will have to pay three months from today to completely pay off the credit card debt.
Section 2.4 - Unknown Time
In an account that has a nominal annual interest rate of i(m), compounded m times per year, a deposit of A( 0 ) will grow to
A(t) = A( 0 )
i(m) m
)mt
by time t. So if A is a target amount for the account total, it will be achieved when
A = A( 0 )
i(m) m
)mt or t = ln(A) − ln(A( 0 )) m ln( 1 + i (m) m ) The account will achieve the multiplication factor of k times the
original deposit when
which does not depend on the original deposit amount, A( 0 ).
Another investment problem, in which we solve for time is when payments of sj are made at times tj , for j = 1 , 2 , · · · , n and we set up an equation of value equating this stream to one payment of (s 1 + s 2 + · · · + sn) at time t. the objective being to find t. The equation of value is then
(s 1 + s 2 + · · · + sn)νt^ = s 1 νt^1 + s 2 νt^2 + · · · + snνtn^.
Of course this can be solved for t using logarithms, provided the right side is not too complicated to compute,
t 0 =
ln(s 1 νt^1 + s 2 νt^2 + · · · + snνtn^ ) − ln(s 1 + s 2 + · · · + sn) ln(ν)
An approximate solution is found with the Method of Equated Time
weighted mean
Example: The present value of a payment of $100 at the end of n years plus another $100 at the end of 2n years is $90. If i = .05 is the annual effective interest rate, determine n.
where ν =
Let x = νn, then the above equation is The solution is then
x =
=. 57238 (+ root needed).
Thus
n = ln(x) ln(ν)
ln(. 57238 ) − ln( 1. 05 )
= 11. 436 years.
Exercise 2-9: A payment of n is made at the end of n years, 2n at the end of 2n years, · · · , n^2 at the end of n^2 years. Find the value of t by the method of equated time.
Section 2.5 - Unknown Rate of Interest
Suppose the principal invested, A( 0 ), accumulates to A(t) at the end of t years. If the interest is compounded m times per year, the solution for the nominal annual interest rate is found via
A( 0 )
i(m) m
)mt = A(t),
which produces
The solution for the annual effective interest rate is
i =
( (^) A(t) A( 0 )
) (^1) t − 1.
Example: Suppose $1000 is invested for 2.5 years and grows in that time to $1220 in an account that is convertible semiannually. What are the effective and nominal interest rates of this investment?
Effective:
i =
Nominal:
i(^2 )^ = 2
[( 1220
]
Exercise2-18: The sum of the accumulated value of 1 at the end of three years at a certain effective interest rate of i and the present value of 1 to be paid at the end of three years at an effective rate of discount that is numerically equal to i is 2.0096. Find i.
Exercise 2-22: A bill for $100 is purchased for $96 three months before it is due. Find (a) the nominal rate of discount convertible quarterly earned by the purchaser and (b) the annual effective rate of interest earned by the purchaser.
